a) By plugging $u=\phi(x-vt)$ into equations (16) and (17) on the problems page, we obtain:\begin{equation}v^{2}\phi''-c^{2}\phi''+m^{2}\phi=0\end{equation}\begin{equation}v^{2}\phi''-c^{2}\phi''-m^{2}\phi=0\end{equation}for (16) and (17), respectively.Proceeding to solve (1) using methods of linear ODEs yields:\begin{equation}\phi''=-\frac{m^{2}}{v^{2}-c^{2}}\phi\Rightarrow \phi(z)=a\sin \left(\frac{m}{\sqrt{v^{2}-c^{2}}}z\right)+b\cos\left(\frac{m}{\sqrt{v^{2}-c^{2}}}z\right)\end{equation}Where $a,b$ depend on initial conditions. This is assuming that $v^{2} > c^{2}$, which gives us a periodic solution that is bounded, as desired. Proceeding the same way with (2):\begin{equation}\phi''=\frac{m^{2}}{v^{2}-c^{2}}\phi\end{equation}If we assume again that $v^{2} > c^{2}$, we will get:\begin{equation}\large\phi(z)=ae^{\frac{m}{\sqrt{v^{2}-c^{2}}}z}+be^{-\frac{m}{\sqrt{v^{2}-c^{2}}}z}\end{equation}But this solution does not tend to $0$ as $|z|\rightarrow\infty$ unless $a=b=0$, and we do not want the trivial solution. Therefore $v^{2}$ must be less than $c^{2}$. Then:\begin{equation}\phi''=-\frac{m^{2}}{c^{2}-v^{2}}\phi\Rightarrow \phi(z)=a\sin \left(\frac{m}{\sqrt{c^{2}-v^{2}}}z\right)+b\cos\left(\frac{m}{\sqrt{c^{2}-v^{2}}}z\right)\end{equation}Now $u=\phi(x-vt)$ using $\phi(z)$ in (3) and (6) solve the original problems (16) and (17), respectively.

b) Plugging $u=\phi(x-vt)$ into problem (18):\begin{equation}-v\phi'-K\phi'''=0\end{equation}\begin{equation}\phi'''+\frac{v}{K}\phi'=0\end{equation}If $v$ and $K$ have the same sign, then the characteristic equation will have imaginary roots and give a bounded periodic solution:\begin{equation}\phi(z)=a\sin \left(\sqrt{\frac{v}{K}}z\right)+b\cos \left(\sqrt{\frac{v}{K}}z\right)+c\end{equation}Same with problem (19):\begin{equation}-v\phi'-iK\phi''=0\end{equation}\begin{equation}\phi''-\frac{iv}{K}\phi'=0\end{equation}In this case, the solution will be suitable regardless of the value of $v$, and the solution will be complex:\begin{equation}\large\phi(z)=ae^{\frac{iv}{K}z}+b=a\left(\cos \left(\frac{vz}{K}\right)+i\sin \left(\frac{vz}{K}\right)\right)+b\end{equation}Same with problem (20):\begin{equation}v^{2}\phi''+K\phi''''=0\Rightarrow\phi''''+\frac{v^{2}}{K}\phi''=0\end{equation}Here, since $v$ only appears in terms of its square, we do not need to worry about its value. As long as $K>0$, solving the above ODE gives us a suitable solution:\begin{equation}\phi(z)=a\sin \left(\frac{vz}{\sqrt{K}}\right)+b\cos \left(\frac{vz}{\sqrt{K}}\right)+c+d\end{equation}Again, $u=\phi(x-vt)$ using $\phi(z)$ in (9), (12), and (14) solve the original problems (18), (19), and (20), respectively.

You do well but I just explain it simpler. Right--we are looking for solutions $u=\phi (x-vt)$ and get\begin{equation}(v^2-c^2)\phi'' + m^2 \phi=0.\label{A}\end{equation}It is ODE and we are looking for its bounded non-trivial solutions. If $v^2-c^2>0$ then such solutions indeed exist and they are (up to the shift) $A\cos (\omega \xi)$ with $\omega= m/\sqrt{v^2-c^2}$ and they are periodic with period $2\pi /\omega = 2\pi \sqrt{v^2-c^2}/m$.

Conclusion: for each $v:|v|\ge c$ there exist required solutions which are periodic with period $2\pi \sqrt{v^2-c^2}/m$.